Question

In the given figure, length of AD is : (A) a √15 units (B) a² √15 units (c) a/2 V15 units (D) a/3 √15 units​

Answer 1

Answer:

AD=√15a/2

Step-by-step explanation:

h²=p²+b²

(2a)²=p²+(a/2)²

4a²-a²/4=p²

15a²/4=p²

√15a/2=P

AD=√15a/2

Answer 2

Step-by-step explanation:

Given: Figure

To find: Length of AD is :

(A) a √15 units

(B) a² √15 units

(C) a/2 √15 units

(D) a/3 √15 units

Solution:

Tip:In isosceles triangle perpendicular on base bisects the base.

Step 1: Write the dimensions of ∆ ABD.

According to the property of isosceles triangle

$BD= \frac{a}{2}$

$AB= 2a$

Step 2: Apply Pythagoras theorem.

As ∆ABD is right triangle, right angle at D.

Thus,

$AB²=AD²+BD²$

$(2a)²=AD²+\left( \frac{a}{2}\right) ²$

or

$AD² = 4 {a}^{2} - \frac{ {a}^{2} }{4}$

or

$AD² = \frac{16 {a}^{2} - {a}^{2} }{4}$

or

$AD² = \frac{15 {a}^{2} }{4}$

or

$\bf AD = \frac{ \sqrt{15} \: a }{2}\:units$

Final answer:

Length of $AD = \frac{a \sqrt{15} }{2} \: units$

Option C is correct.

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